常见算法 - 按大小合并多个有序链表

leetcode(23)：

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

思路：可以想到可以两两合并链表，循环一次，再两两合并(leetcode21)，直到合并成一个。那么就可以想到用二分法，然后两两合并。当然如果可以，也可以用其他有规律的选择两两合并。

public class L23MergekSortedLists {
	 public  static ListNode mergeKLists(ListNode[] lists) {
	        
		 if(lists.length == 0){
			 return null;
		 }else if(lists.length == 1){
			 return lists[0];
		 }
		 int length = lists.length;
		 
		 while(length > 1){
			 int mid = (length+1)/2;
			 for (int i = 0; i < length/2; i++) {
				lists[i] = mergeTwoList(lists[i],lists[mid+1]);
			 }
			 length = mid;
		 }
		 return lists[0];
	 }

	private static ListNode mergeTwoList(ListNode listNode, ListNode listNode2) {
		
		if(listNode == null && listNode2 == null){
			return null;
		}else if(listNode == null){
			return listNode2;
		}else if(listNode2 == null){
			return listNode;
		}
		ListNode head = new ListNode(0);
		ListNode dummyHead = head;
		while(listNode != null && listNode2 != null){
			if(listNode.val >= listNode2.val){
				head.next = listNode;
				listNode = listNode.next;
			}else {
				head.next = listNode2;
				listNode2 = listNode2.next;
			}
			head = head.next;
		}
		if(listNode != null){
			head.next = listNode;
		}
		if(listNode2 != null){
			head.next = listNode2;
		}
		return dummyHead.next;
	}
}